Question

An electron is released from the negative plate of a parallel plate capacitor with an internal field strength of 2.5×104 V/m and a 1.5 mm spacing. (a) What is the speed of the electron once it reaches the positive plate? (b) What is the electron’s acceleration?

Answer 1

Part A.

Using Work-energy theorem:

W = dKE

W = Work-done by potential difference = q*dV

dKE = KEf - KEi

KEi = 0, since initial speed of electron is zero,

kEf = (1/2)*m*V^2

So,

q*dV = (1/2)*m*V^2 - 0

V = sqrt (2*q*dV/m)

dV = Potential difference between plates = E*d = 2.5*10^4*1.5*10^-3 = 37.5 V

Using given values:

V = sqrt (2*1.6*10^-19*37.5/(9.1*10^-31))

V = Speed of electron on the positive plate = 3.63*10^-6 m/s

Part B.

Using 3rd kinematic equation:

V^2 = U^2 + 2*a*d

a = acceleration of electron = (V^2 - U^2)/(2*d)

a = [(3.63*10^6)^2 - 0^2]/(2*1.5*10^-3)

a = 4.39*10^15 m/s^2

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